How can the rationalized numerator be simplified?

[r0bHadz]

Homework Statement

$\frac {\sqrt{2x+3}+1}4$

Homework Equations

The Attempt at a Solution

$\frac {\sqrt{2x+3}+1}4$ * $\frac {\sqrt{2x+3}-1}{\sqrt{2x+3}-1}$ = $\frac {2x+3-1}{4\sqrt{2x+3}-4}$ = $\frac {2x+2}{4\sqrt{2x+3}-4}$ = $\frac {2(x+1)}{2(2\sqrt{2x+3}-2)}$ = $\frac {x+1}{2\sqrt{2x+3} -2}$

but doc. Lang is telling me the answer is $\frac {x+1}{2(\sqrt{2x+3}-2)}$

How did he come to this result??

 

[SammyS]

Homework Statement

$\frac {\sqrt{2x+3}+1}4$

Homework Equations

The Attempt at a Solution

$\frac {\sqrt{2x+3}+1}4$ * $\frac {\sqrt{2x+3}-1}{\sqrt{2x+3}-1}$ = $\frac {2x+3-1}{4\sqrt{2x+3}-4}$ = $\frac {2x+2}{4\sqrt{2x+3}-4}$ = $\frac {2(x+1)}{2(2\sqrt{2x+3}-2)}$ = $\frac {x+1}{2\sqrt{2x+3} -2}$

but doc. Lang is telling me the answer is $\frac {x+1}{2(\sqrt{2x+3}-2)}$

How did he come to this result??

Yes, you are correct.

 

[r0bHadz]

Yes, you are correct.

Thank you no more discussion is needed. I'm going to assume Lang didn't intend for there to be "(" after the first two in the denominator

 

[SammyS]

Thank you no more discussion is needed. I'm going to assume Lang didn't intend for there to be "(" after the first two in the denominator

... or he intended the denominator to be: $\ 2(\sqrt{2x+3\,} -1)\,.$

 

[r0bHadz]

... or he intended the denominator to be: $\ 2(\sqrt{2x+3\,} -1)\,.$

True that.